Is it trivial to say $$\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k},$$
considering the fact that we know $$\mathop {\lim }\limits_{n \to \infty } {(1 + {1 \over n})^n} = e?$$
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4$\begingroup$ The first limit is $e^k$. $\endgroup$ – user63181 Nov 15 '13 at 15:30
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1$\begingroup$ can you tell me why? $\endgroup$ – captain dragon Nov 15 '13 at 15:33
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3$\begingroup$ $\lim\limits_{n \to \infty } {\left(1 + \frac{k}{n}\right)^n} = \lim\limits_{n \to \infty } {(( 1 + \frac{1}{\frac{n}{k}})^{\frac{n}{k}}})^k=e^k$ $\endgroup$ – user35603 Nov 15 '13 at 15:36
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1$\begingroup$ oh, got it. please write it as an answer $\endgroup$ – captain dragon Nov 15 '13 at 15:37
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$\begingroup$ please, fix the OP $e\to e^k$ $\endgroup$ – Ilya Nov 15 '13 at 15:49
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Since $\frac kn = \frac 1 {\frac nk}$ and $n = \frac nk \cdot k$ we have $$\lim_{n\to \infty} (1+\frac kn)^n = \lim_{n\to\infty} \left( (1 + \frac{1}{\frac nk})^{\frac nk} \right)^k = \left ( \lim_{u\to\infty} (1+\frac 1u)^u \right)^k = e^k$$ If $k<0$, you have $$(1+\frac 1u)^{-u} = \left((1+\frac 1u)^u\right)^{-1} \to e^{-1} \qquad (u\to\infty)$$
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3$\begingroup$ That's fine for $k > 0$. But the statement is actually true for all $k$. $\endgroup$ – Robert Israel Nov 15 '13 at 15:42
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$\begingroup$ @RobertIsrael Your assertion is correct, thanks.# $\endgroup$ – AlexR Nov 15 '13 at 15:45
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Well, by (one) definition of $e$, we have $$e=\lim_{t\to\infty}\left(1+\frac1t\right)^t.$$ Now, in particular, if $k>0$, then letting $t=\frac nk,$ we find that $t\to\infty$ precisely as $n\to\infty,$ so that $$\begin{align}\lim_{n\to\infty}\left(1+\frac kn\right)^n &= \lim_{n\to\infty}\left(1+\frac kn\right)^{\frac nk\cdot k}\\ &= \lim_{t\to\infty}\left(1+\frac1t\right)^{tk}\\ &= \lim_{t\to\infty}\left(\left(1+\frac1t\right)^t\right)^k\\ &= \left(\lim_{t\to\infty}\left(1+\frac1t\right)^t\right)^k\\ &= e^k\end{align}$$ The $k=0$ case is trivial. To deal with the $k<0$ case, observe that $$1+\frac kn=\frac{n+k}{n}=\left(\frac{n}{n+k}\right)^{-1}=\left(1+\frac{-k}{n+k}\right)^{-1},$$ so letting $t=\frac{n+k}{-k},$ we once again have $t\to\infty$ as $n\to\infty,$ and so $$\begin{align}\lim_{n\to\infty}\left(1+\frac kn\right)^n &= \lim_{n\to\infty}\left(1+\frac{-k}{n+k}\right)^{-n}\\ &= \lim_{n\to\infty}\left(1+\frac{-k}{n+k}\right)^{-(n+k)+k}\\ &= \lim_{n\to\infty}\left(1+\frac{-k}{n+k}\right)^{-(n+k)}\left(1+\frac{-k}{n+k}\right)^k\\ &= \lim_{n\to\infty}\left(1+\frac{-k}{n+k}\right)^{-(n+k)}\lim_{n\to\infty}\left(1+\frac{-k}{n+k}\right)^k\\ &= \lim_{n\to\infty}\left(1+\frac kn\right)^{\frac{n+k}{-k}\cdot k}\\ &= \lim_{t\to\infty}\left(1+\frac1t\right)^{tk}\\ &= \lim_{t\to\infty}\left(\left(1+\frac1t\right)^t\right)^k\\ &= \left(\lim_{t\to\infty}\left(1+\frac1t\right)^t\right)^k\\ &= e^k\end{align}$$
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By $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e,n\in\mathbb{N}$, we have $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e,\quad x\in \mathbb{R}$$
